15.8%20Ex%2041%20-%2042

15.8%20Ex%2041%20-%2042 - 798 C H A P T E R 15 D I F F E R...

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798 CHAPTER 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) (c) The equality obtained in part (b) implies that λ is the rate of change in the maximum value of f ( x , y ) , subject to the constraint g ( x , y ) = c , with respect to c . 41. Let B > 0. Show that the maximum of f ( x 1 ,... , x n ) = x 1 x 2 ··· x n subject to the constraints x 1 +···+ x n = B and x j 0for j = 1 n occurs for x 1 =···= x n = B / n . Use this to conclude that ( a 1 a 2 a n ) 1 / n a 1 a n n for all positive numbers a 1 a n . SOLUTION We Frst notice that the constraints x 1 x n = B and x j j = 1 n deFne a closed and bounded set in the n th dimensional space, hence f (continuous, as a polynomial) has extreme values on this set. The minimum value zero occurs where one of the coordinates is zero (for example, for n = 2 the constraint x 1 + x 2 = B , x 1 0, x 2 0 is a triangle in the Frst quadrant). We need to maximize the function f ( x 1 x n ) = x 1 x 2 x n subject to the constraints g ( x 1 x n ) = x 1 x n B = 0, x j 0, j = 1 n . Step 1. Write out the Lagrange Equations. The gradient vectors are f = - x 2 x 3 x n , x 1 x 3 x n x 1 x 2 x n 1 ® g = h 1 , 1 1 i The Lagrange Condition f = g yields the following equations: x 2 x 3 x n = x 1 x 3 x n = x 1 x 2 x n 1 = Step 2. Solving for x 1 , x 2 x n using the constraint. The Lagrange equations imply the following equations: x 2 x 3 x n = x 1 x 2 x n 1 x 1 x 3 x n = x 1 x 2 x n 1 x 1 x 2 x 4 x n = x 1 x 2 x n 1 .
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15.8%20Ex%2041%20-%2042 - 798 C H A P T E R 15 D I F F E R...

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